If z and omega are two non-zero complex numbe | vedclass

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(D) Given $|z\omega| = 1$,which implies $|z||\omega| = 1$ ... $(i)$Also,$	ext{arg}(z) - 	ext{arg}(\omega) = \frac{\pi}{2}$,which implies $	ext{arg}

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$-i$

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(D) Given $|z\omega| = 1$,which implies $|z||\omega| = 1$ ... $(i)$Also,$	ext{arg}(z) - 	ext{arg}(\omega) = \frac{\pi}{2}$,which implies $	ext{arg}(\frac{z}{\omega}) = \frac{\pi}{2}$.This means $\frac{z}{\omega} = |\frac{z}{\omega}| e^{i\pi/2} = |\frac{z}{\omega}| i$. Since $|\frac{z}{\omega}| = \frac{|z|}{|\omega|}$,we have $\frac{z}{\omega} = \frac{|z|}{|\omega|} i$ ... $(ii)$From $(i)$,$|z| = \frac{1}{|\omega|}$. Substituting this into $(ii)$,we get $\frac{z}{\omega} = \frac{1}{|\omega|^2} i$.However,a simpler approach: $\frac{z}{\omega} = i \implies z = i\omega$.Taking the conjugate,$\bar{z} = \bar{i}\bar{\omega} = -i\bar{\omega}$.Then $\bar{z}\omega = (-i\bar{\omega})\omega = -i(\bar{\omega}\omega) = -i|\omega|^2$.Since $|z\omega| = 1$ and $\frac{z}{\omega} = i$,we have $|z| = |i\omega| = |\omega|$,so $|\omega|^2 = |z||\omega| = 1$.Therefore,$\bar{z}\omega = -i(1) = -i$.

If $z$ and $\omega$ are two non-zero complex numbers such that $|z\omega| = 1$ and $\text{arg}(z) - \text{arg}(\omega) = \frac{\pi}{2}$,then $\bar{z}\omega$ is equal to

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